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 A226593 Period of longest sequence generated by recursive permutation of permutation pairs of length n. 1

%I

%S 1,3,8,18,96,216

%N Period of longest sequence generated by recursive permutation of permutation pairs of length n.

%C The n! permutations (p) of the numbers 1,2,3..n may be paired (allowing duplication) in n!^2 ways. Let p’ represent a permutation of the identity 123..n: then in p x p’ = p’’, p’ will (by x) permute p (in the same way the identity was permuted) to p’’. For example, 2143 x 4321 = 3412. Iterating, 4321 x 3412 = 2143 for a period of 3. If p = p’, this recursive process gives the Pisano periods. For most other pairings the periods are of different lengths. The sequence gives the longest period that p x p’ generates for any p of length n.

%H Russell Walsmith, <a href="/A226593/a226593_1.pdf">An investigation of recursive sequences based on pairs of permutations of length n.</a>

%e For n = 4: 3142 x 2341 = 1423; 2341 x 1423 = 2134... the sequence thus generated is of period = 18.

%o (PARI) period(a,b)=my(n=matsize(a)[2], v=vector(n), aa=vector(n,i,a[i]), bb=vector(n,i,b[i]), id, nsteps); while(id!=n, for(i=1,n, v[i]=a[b[i]]); id=sum(i=1,n, b[i]==aa[i] && v[i]==bb[i]); for(i=1,n, a[i]=b[i]; b[i]=v[i]); nsteps++); nsteps

%o a(n)=my(a,b,m,p); for(k=1,n!, a=numtoperm(n,k); for(l=1,n!, b=numtoperm(n,l); p=period(a,b); if(p>m,m=p))); m \\ _Ralf Stephan_, Aug 13 2013

%Y Cf. A001175 (Pisano periods: period of Fibonacci numbers (A000045) mod n).

%Y Cf. A106291 (period of the Lucas sequence (A000032) mod n).

%K nonn,hard,more

%O 1,2

%A _Russell Walsmith_, Jun 13 2013

%E a(6) from _Ralf Stephan_, Aug 13 2013

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Last modified June 23 07:00 EDT 2021. Contains 345395 sequences. (Running on oeis4.)